As snipez90 said, if you let I=the original integral, then if you do integration by parts once more (so the sin turns back into a cos), you should get (a bunch of stuff)-I. So I=(a bunch of stuff)-I. Solve for I.

As snipez90 said, if you let I=the original integral, then if you do integration by parts once more (so the sin turns back into a cos), you should get (a bunch of stuff)-I. So I=(a bunch of stuff)-I. Solve for I.

I didn't check the details of the integration, but yes, that's the general idea. Now call I the integral of cos(x)*exp(-inx). It occurs on both sides of your equation. Solve for it.

changed my post again,

If my integration and if my rearranging of the equation is what you are refering too. Do I then as a final step covert the sum of the two integrals into one sum which then results in the right-hand side of the equation?

If my integration and if my rearranging of the equation is what you are refering too. Do I then as a final step covert the sum of the two integrals into one sum which then results in the right-hand side of the equation?

I'm not quite sure what you mean there, but what I meant is to write 'I' for the integral. So the left side of your equation is I/(2pi)+I*i/n^2. Factor the I out, like I*(1/(2pi)+i/n^2). Now solve for I.

Your result for [itex]c_n[/itex] is correct, but the work you've shown on the integration leads me to believe you might still be a little lost.... If so ,first redo your integration by parts without forgetting the factor of 1/2pi...you should get:

Your result for [itex]c_n[/itex] is correct, but the work you've shown on the integration leads me to believe you might still be a little lost.... If so ,first redo your integration by parts without forgetting the factor of 1/2pi...you should get: